# Euler-Lagrange equation

1. Dec 26, 2007

### neworder1

1. The problem statement, all variables and given/known data

Let $$P$$ be a rectangle , $$f_{0} : \partial P \rightarrow R)$$ continuous and Lipschitz, $$C_{0} = \{ f \in C^{2}(P) : f=f_{0} \ on \ \partial P \}$$. and finally $$S : C_{0} \rightarrow R$$ a functional:

$$S(f) = \int^b_a (\int^d_c (\frac{\partial f}{\partial x})^{2}\,dy)\,dx + \int^d_c (\int^a_b (\frac{\partial f}{\partial y})^{2}\,dx)\,dy$$.

Write Euler-Lagrange equation for S.

2. Relevant equations

3. The attempt at a solution

I tried writing: $$S(f) = \int^b_a (\int^d_c (\frac{\partial f}{\partial x})^{2} + (\frac{\partial f}{\partial y})^{2}\,dy)\,dx$$, so the proper Lagrangian would be $$L(x) = \int^d_c (\frac{\partial f}{\partial x})^{2} + (\frac{\partial f}{\partial y})^{2}\,dy$$.

Then the Euler-Lagrange equation should be $$\frac{d}{dx}\frac{\partial L}{\partial f^{'}_{x}} = 0 \leftrightarrow \frac{d}{dx}\frac{\partial }{\partial f^{'}_{x}}\int^d_c (\frac{\partial f}{\partial x})^{2}\,dy = 0$$, ($$L^{'} = \int^d_c (\frac{\partial f}{\partial x})^{2}\,dy$$) and now since $$\frac{dL^{'}}{dx} = \frac{\partial L^{'}}{\partial f_{x}^{'}}\frac{\partial f_{x}^{'}}{\partial x}$$, we can rewrite that as $$\frac{d}{dx}\frac{\frac{d}{dx}\int^d_c (\frac{\partial f}{\partial x})^{2}\,dy}{f_{xx}^{''}} = 0 \leftrightarrow \frac{d}{dx}\frac{\int^d_c \frac{\partial}{\partial x}(\frac{\partial f}{\partial x})^{2}\,dy}{f_{xx}^{''}} = 0 \leftrightarrow \frac{d}{dx}\frac{\int^d_c 2f_{x}^{'}f_{xx}^{''}\,dy}{f_{xx}^{''}} = 0$$. but what then?

Last edited: Dec 26, 2007
2. Dec 26, 2007

### Dick

You are making this too hard. Let's call $$\frac{\partial f}{\partial x}=f_x$$ and $$\frac{\partial f}{\partial y}=f_x$$. Then the form of the Euler-Lagrange equations for two independent variables is $$\frac{\partial L}{\partial f}-\frac{\partial}{\partial x} \frac{\partial L}{\partial f_x}-\frac{\partial}{\partial y}\frac{\partial L}{\partial f_y}=0$$ where $$L=(f_x)^2+(f_y)^2$$.